$$f\left( {\frac{\pi }{2}} \right) = 3.$$   Since $$f\left( x \right)$$  is continuous at $$x = \frac{\pi }{2}$$
$$\eqalign{ & \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{2}} = \left( {\frac{{k\,\cos \,x}}{{\pi - 2x}}} \right) = f\left( {\frac{\pi }{2}} \right) \cr & \Rightarrow \frac{k}{2} = 3 \cr & \Rightarrow k = 6 \cr} $$

Since, $$f\left( x \right)$$  is continuous, then
$$\eqalign{ & \mathop {\lim }\limits_{x \to \frac{\pi }{4}} f\left( x \right) = f\left( {\frac{\pi }{4}} \right) \cr & \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sqrt 2 \,\cos \,x - 1}}{{\cot \,x - 1}} = k \cr} $$
Now by L-hospital’s rule
$$\eqalign{ & \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sqrt 2 \,\sin \,x}}{{{\text{cose}}{{\text{c}}^2}\,x}} = k \cr & \Rightarrow \frac{{\sqrt 2 \left( {\frac{1}{{\sqrt 2 }}} \right)}}{{{{\left( {\sqrt 2 } \right)}^2}}} = k \cr & \Rightarrow k = \frac{1}{2} \cr} $$

When $$x$$ is not an integer, both the functions $$\left[ x \right]$$ and $$\cos \left( {\frac{{2x - 1}}{2}} \right)\pi $$    are continuous.
$$\therefore \,f\left( x \right)$$   continuous on all non integral points.
For $$x = n \in I$$
$$\eqalign{ & \mathop {\lim }\limits_{x \to {n^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi \cr & = \left( {n - 1} \right)\cos \left( {\frac{{2n - 1}}{2}} \right)\pi = 0 \cr & \mathop {\lim }\limits_{x \to {n^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {n^ + }} \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi \cr & = n\cos \left( {\frac{{2n - 1}}{2}} \right)\pi = 0 \cr & {\text{Also }}f\left( n \right) = n\,\cos \frac{{\left( {2n - 1} \right)\pi }}{2} = 0 \cr} $$
$$\therefore \,f$$  is continuous at all integral pts. as well.
Thus, $$f$$ is continuous everywhere.

Apply $$\mathop {\lim }\limits_{h \to 0} \frac{{1 - \sin \left( {\frac{\pi }{2} + h} \right)}}{{\sin \,2\left( {\frac{\pi }{2} + h} \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \sin \left( {\frac{\pi }{2} - h} \right)}}{{\sin \,2\left( {\frac{\pi }{2} - h} \right)}} = f\left( {\frac{\pi }{2}} \right)$$

The function $$\log \left| x \right|$$  is not defined at $$x = 0.$$
So, $$x = 0$$  is a point of discontinuity .
Also, for $$f\left( x \right)$$  to be defined ; $$\log \left| x \right| \ne 0 \Rightarrow x \ne \pm 1$$
Hence, $$0,\,1,\, - 1$$   are three points of discontinuity.

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0 + 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \left[ {{{\tan }^2}h} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,0 = 0 \cr & \mathop {\lim }\limits_{x \to 0 - 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 - h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \left[ {{{\tan }^2}\left( { - h} \right)} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,0 = 0\,;\,f\left( 0 \right) = 0 \cr & \therefore \mathop {\lim }\limits_{h \to 0} f\left( x \right) = f\left( 0 \right) = 0 \cr & \therefore f\left( x \right){\text{ continuous at }}x = 0 \cr & \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\tan }^2}h} \right] - 0}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{0}{h} = \mathop {\lim }\limits_{h \to 0} \,0 = 0 \cr & \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\tan }^2}\left( { - h} \right)} \right] - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{0}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \,0 = 0 \cr & {\text{So, }}f'\left( 0 \right) = 0 \cr} $$

Using Lagrange's Mean Value Theorem
Let $$f\left( x \right)$$  be a function defined on $$\left[ {a,\,b} \right]$$
Then
$$\eqalign{ & f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}.....({\text{i}}) \cr & c\, \in \left[ {a,\,b} \right] \cr & \therefore \,{\text{Given }}f\left( x \right) = {\log _e}x \cr & \therefore \,f'\left( x \right) = \frac{1}{x} \cr & \therefore {\text{ equation (i) become }}\frac{1}{c} = \frac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}} \cr & \Rightarrow \frac{1}{c} = \frac{{{{\log }_e}3 - {{\log }_e}1}}{2} = \frac{{{{\log }_e}3}}{2} \cr & \Rightarrow c = \frac{2}{{{{\log }_e}3}} \cr & \Rightarrow c = 2\,{\log _3}e \cr} $$

Let $$f\left( x \right)$$  is continuous at $$x = 1,$$   then
$$\eqalign{ & f\left( {{1^ - }} \right) = f\left( a \right) = f\left( {{1^ + }} \right) \cr & \Rightarrow 5 = a + b......(a) \cr} $$
Let $$f\left( x \right)$$  is continuous at $$x = 3,$$   then
$$\eqalign{ & f\left( {{3^ - }} \right) = f\left( c \right) = f\left( {{3^ + }} \right) \cr & \Rightarrow a + 3b = b + 15 \cr & \Rightarrow a + 2b = 15......(b) \cr} $$
Solving (a) & (b) we get $$b= 10, \,\,a =-5$$
Now $$f\left( x \right)$$  is continuous at $$x = 5,$$   then
$$\eqalign{ & f\left( {{5^ - }} \right) = f\left( 5 \right) = f\left( {{5^ + }} \right) \cr & \Rightarrow b + 25 = 30 \cr} $$
Which is not satisfied by $$a =-5$$   and $$b= 10.$$
Hence, $$f\left( x \right)$$  is not continuous for any values of $$a$$ and $$b$$

\[\begin{array}{l} f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {\left( {{{\cos }^2}x} \right)^n}\\ = \left\{ \begin{array}{l} 0,\,\,\,0 \le {\cos ^2} x < 1\\ 1,\,\,\,\,\,\,{\cos ^2}x = 1 \end{array} \right.\\ = \left\{ \begin{array}{l} 0,\,\,\,\,x \ne n\pi ,\,n\, \in \,I\\ 1,\,\,\,\,x = n\pi ,\,n\, \in \,I \end{array} \right. \end{array}\]
Hence, $$f\left( x \right)$$  is discontinuous when $$x = n\pi ,\,n\, \in \,I$$
For this values of $$\theta ,\,\sin \,\theta = 0.$$

Let \[h\left( x \right)\left| \begin{array}{l} f\left( a \right)\,\,\,\,\,f\left( x \right)\\ g\left( a \right)\,\,\,\,\,g\left( x \right) \end{array} \right| = f\left( a \right)g\left( x \right) - g\left( a \right)f\left( x \right)\]
Then, \[h'\left( x \right) = f\left( a \right)g'\left( x \right) - g\left( a \right)f'\left( x \right) = \left| \begin{array}{l} f\left( a \right)\,\,\,\,\,f'\left( x \right)\\ g\left( a \right)\,\,\,\,\,g'\left( x \right) \end{array} \right|\]
Since, $$f\left( x \right)$$  and $$g\left( x \right)$$  are continuous in $$\left[ {a,\,b} \right]$$  and differentiable in $$\left( {a,\,b} \right),$$  therefore $$h\left( x \right)$$  is also continuous $$\left[ {a,\,b} \right]$$  in and differentiable in $$\left( {a,\,b} \right).$$
So, by mean value theorem, there exists at least one real number $$c,\,a < c < b$$   for which $$h'\left( c \right) = \frac{{h\left( b \right) - h\left( a \right)}}{{b - a}},$$
$$\therefore \,h\left( b \right) - h\left( a \right) = \left( {b - a} \right)h'\left( c \right).....\left( {\text{i}} \right)$$
Here, \[h\left( a \right) = \left| \begin{array}{l} f\left( a \right)\,\,\,\,\,f\left( a \right)\\ g\left( a \right)\,\,\,\,\,g\left( a \right) \end{array} \right| = 0,\,\,h\left( b \right) = \left| \begin{array}{l} f\left( a \right)\,\,\,\,\,f\left( b \right)\\ g\left( a \right)\,\,\,\,\,g\left( b \right) \end{array} \right|\]

$$\therefore $$  From equation \[\left( {\rm{i}} \right),\,\left| \begin{array}{l} f\left( a \right)\,\,\,\,\,f\left( b \right)\\ g\left( a \right)\,\,\,\,\,g\left( b \right) \end{array} \right| = \left( {b - a} \right),\,\,h'\left( c \right) = \left( {b - a} \right)\left| \begin{array}{l} f\left( a \right)\,\,\,\,\,f'\left( c \right)\\ g\left( a \right)\,\,\,\,\,g'\left( c \right) \end{array} \right|\]